MESSAGE
| DATE | 2016-11-26 |
| FROM | Ruben Safir
|
| SUBJECT | Subject: [Learn] operator<<() overloading details and friend
|
From learn-bounces-at-nylxs.com Sat Nov 26 18:08:10 2016
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Subject: [Learn] operator<<() overloading details and friend
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* zacklocx (~zacklocx-at-114.111.167.113) has joined ##c++
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sorry -- should be `[const] Input` for copy by value
mrbrklyn: Did you mean operator<<()?
yes
operator<<()
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mrbrklyn: And did you mean operator<<(std::ostream&, const T &)?
* sid_fules (~sid_fules-at-79-75-34-79.dynamic.dsl.as9105.com) has joined ##c++
i.e. the "streaming" operator, rather than the bitshift
operator?
std::ostream& operator << ( std::ostream &output,
state const & p )
yes
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it doesn't need to be a friend unless it needs special access
You may want to print out some internal state. To do so, you
may require access to private or protected properties.
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why not just make it part of the class like other operators?
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One workaround would be to define a public member function
void print(std::ostream &), though; then you op<< can just call that
(p.print(output); return output; in your example) instead of having to
be declared a friend.
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mrbrklyn: Because if you make an op<< a member function of a
class, its left hand side operator will be an instance of the class it
is defined in.
mrbrklyn: you can't make it a member because that would mean
the left side is of the class type
So you'd need to define it inside of std::ostream - which
obviously is a bad idea.
its left side must be ostream&
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because it can be used as an l value?
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I only seem to ever sort of understand this, and never fully
understand it
it has defied me for years
If you define any operator as a member function, the type
it is defined inside is the left-hand side argument
Just by the definition of member functions
However, arbitrary user-defined types want to overload the
stream operator to use ostream
They can't all be inside of the ostream class because
ostream doesn't know about those types
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has joined ##c++
mrbrklyn: You can define them as a friend within the class,
though.
Or as just a free function
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struct my_type { friend std::ostream
&operator<<(std::ostream &os, const my_type &obj) { return os <<
obj.value; } private: int value; };
^ is that a free function?
It is.
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It is shorthand for: struct my_type { friend std::ostream
&operator<<(std::ostream &os, const my_type &obj); private: int value;
}; std::ostream &operator<<(std::ostream &os, const my_type &obj) {
return os << obj.value; }
Plus I'd assume it might also imply op<< being inline, just
like defining member functions within the class definition.
Not 100% sure about that, though.
friend op<< is a code smell, because if op<< can't be
implemented with the public interface, it means that you may have an
object x with a property p where the only way to read x's p is to first
print x and then parse what was printed
which is silly, because in such a case there should normally be
a way to just get x's p without going through serialization
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